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3 years ago

If the pressure of a gas is 2.000 × 104 kPa, what is its pressure in atm?

Chemistry

2 answers:

lisov135 [29]3 years ago

7 0

Answer:

197200atm

Explanation:

There are different units to express pressure of a gas, some of them are atmosphere, Bar, Kilopascal, Pascal, torr among other.

In this exercise we need to know the equivalence between atm and Kpa.

if, 1kp=0,00986atm

then,

(2,000 x104kPa)*(0,00986atm)/1Kpa= 0,00986 atm

luda_lava [24]3 years ago

4 0

<u>Answer:</u>

<em>197.4 atm is the Answer.</em>

<u>Explanation:</u>

Pressure can be expressed in unit of torr, mm Hg, psi, atm, kPa etc

So Here to convert atm to kPa the conversion factor is either $\frac {1atm}{101.325kPa}$ or $\frac {101.325kPa}{1atm}$

We need the answer in atm so kPa should get cancel and it should be in the denominator

$2.000 \times 10^4 kPa\times \frac {1atm}{101.325kPa}$

=197.4 atm is the Answer.

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Explanation: so basically rusting happens when a metal comes in contact with moisture and air, at a brown, dirty metal coat forms on it.

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2 years ago

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3 years ago

How are rain hail and snow formed

zavuch27 [327]

When cloud particles become too heavy to remain suspended in the air, they fall to the earth as precipitation. Precipitation occurs in a variety of forms; hail, rain, freezing rain, sleet or snow.
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Hope this helped :)
Have a great day

7 0

3 years ago

Calculate Delta H in KJ for the following reactions using heats of formation:

lozanna [386]

Answer:

$\Delta H\textdegree = -2856.8\;\text{kJ}$ per mole reaction.

$\Delta H\textdegree = -22.3\;\text{kJ}$ per mole reaction.

Explanation:

What is the standard enthalpy of formation $\Delta H_f\textdegree{}$ of a substance? $\Delta H_f\textdegree{}$ the enthalpy change when one mole of the substance is formed from the most stable allotrope of its elements under standard conditions.

Naturally, $\Delta H_f\textdegree{} = 0$ for the most stable allotrope of each element under standard conditions. For example, oxygen $\text{O}_2$ (not ozone $\text{O}_3$ ) is the most stable allotrope of oxygen. Also, under STP $\text{O}_2$ is a gas. Forming $\text{O}_2\;(g)$ from itself does not involve any chemical or physical change. As a result, $\Delta H_f\textdegree{} = 0$ for $\text{O}_2\;(g)$ .

Look up standard enthalpy of formation $\Delta H_f\textdegree{}$ data for the rest of the species. In case one or more values are not available from your school, here are the published ones. Note the state symbols of the compounds (water/steam $\text{H}_2\text{O}$ in particular) and the sign of the enthalpy changes.

$\text{C}_2\text{H}_6\;(g)$ : $-84.0\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{CO}_2\;(g)$ : $-393.5\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{H}_2\text{O}\;{\bf (g)}$ : $-241.8\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{PbO}\;(s)$ : $-217.9\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{PbO}_2\;(s)$ : $-276.6\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{Pb}_3\text{O}_4\;(s)$ : $-734.7\;\text{kJ}\cdot\text{mol}^{-1}$

How to calculate the enthalpy change of a reaction $\Delta H_\text{rxn}$ (or simply $\Delta H$ from enthalpies of formation?

Multiply the enthalpy of formation of each product by its coefficient in the equation.
Find the sum of these values. Label the sum $\Sigma (n\cdot \Delta_f(\text{Reactants}))$ to show that this value takes the coefficients into account.
Multiply the enthalpy of formation of each reactant by its coefficient in the equation.
Find the sum of these values. Label the sum $\Sigma (n\cdot \Delta_f(\text{Products}))$ to show that this value takes the coefficient into account.
Change = Final - Initial. So is the case with enthalpy changes. $\Delta H_\text{rxn} = \Sigma (n\cdot \Delta_f(\textbf{Products})) - \Sigma (n\cdot \Delta_f(\textbf{Reactants}))$ .

For the first reaction:

$\Sigma (n\cdot \Delta_f(\text{Reactants})) = 4\times (-393.5) + 6\times (-241.8) = -3024.8\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\Sigma (n\cdot \Delta_f(\text{Products})) = 2\times (-84.0) + 7\times 0 = -168.0\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\begin{aligned}\Delta H_\text{rxn} &= \Sigma (n\cdot \Delta_f(\textbf{Products})) - \Sigma (n\cdot \Delta_f(\textbf{Reactants}))\\ &= (-3024.8\;\text{kJ}\cdot\text{mol}^{-1}) - (-168.0\;\text{kJ}\cdot\text{mol}^{-1})\\ &= -2856.8\;\text{kJ}\cdot\text{mol}^{-1} \end{aligned}$ .

Try these steps for the second reaction:

$\Delta H_\text{rxn} = -22.3\;\text{kJ}\cdot\text{mol}^{-1}$ .

6 0

4 years ago

Is Newton's statement a law or a theory? It is a theory because it developed and changed over time as new technology provided ne

nika2105 [10]

Answer:

It is a law because it is a statement of fact that does not need any more experimental support.

Explanation:

A hypothesis is an acceptable assumption that is formulated through the collection of information and data and that attempts to explain some phenomenon. In other words, a hypothesis is a conjecture with which one tries to explain an observed fact. The formulation of a hypothesis can be derived in two different ways: by generalization, from a certain number of cases or observed phenomena; and, by analogies observed in the occurrence of different phenomena.

A theory is something that attempts to explain some phenomenon and that is accepted as true by the scientific community. In other words, a theory is a set of rules that describe the behavior of a particular system.

A law is a statement that expresses the constant relationships between phenomena of nature, society or the mind. Scientific law can be defined as an invariable relationship of terms. In other words, a law is what is scientifically proven to be true, it is universally applicable.

So scientific theories and scientific laws are based on hypotheses. But a theory is the explanation of an observed phenomenon, while a scientific law is the description of an observed phenomenon.

The movements of bodies in space are governed by Newton's Laws. These are laws why they are demonstrated by mathematical formulas and Universal application.

Then, the correct option is <u><em>"It is a law because it is a statement of fact that does not need any more experimental support."</em></u>

5 0

4 years ago